NECO 2023 · Paper 1 · Q27

A candidate is asked to draw the graph of y=x2+6x−27y = x^2 + 6x - 27 and a linear graph on the same axes such that their intersections give the solutions of x2+5x−29=0x^2 + 5x - 29 = 0. What is the equation of the linear graph?

Worked solution (try it first)
  1. Setting the curve equal to the line must give the same equation as x2+5x−29=0x^2 + 5x - 29 = 0.
  2. Subtract the equation from the curve: (x2+6x−27)−(x2+5x−29)=x+2(x^2 + 6x - 27) - (x^2 + 5x - 29) = x + 2.
  3. So the line is y=x+2y = x + 2, option E.
  4. Check: x2+6x−27=x+2x^2 + 6x - 27 = x + 2 rearranges to x2+5x−29=0x^2 + 5x - 29 = 0.

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